Finding P(1) Given Polynomial Factors: A Step-by-Step Guide

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Finding P(1) Given Polynomial Factors: A Step-by-Step Guide

Let's dive into this fascinating math problem where we're tasked with finding the value of P(1), given that P(x) is a factor of both T(x) and S(x). Sounds intriguing, right? Well, buckle up, guys, because we're about to break it down in a way that's super easy to understand. We will explore the methods step by step and in detail.

Understanding the Problem

At the heart of our problem lies polynomial factorization. We are given three polynomials:

  • P(x) = x² + bx + c: This is our quadratic polynomial, the one we're trying to understand better. The goal is to find the value of this function when x = 1, i.e., P(1).
  • T(x) = x⁴ + 6x² + 25: A quartic polynomial (degree 4). We know P(x) is a factor of this, which means T(x) can be divided by P(x) without any remainder. This relationship is key to solving our problem.
  • S(x) = 3x⁴ + 4x² + 28x + 5: Another quartic polynomial. Similar to T(x), P(x) is also a factor of S(x). This gives us another crucial piece of information.

Our mission, should we choose to accept it (and we do!), is to leverage the fact that P(x) is a common factor of both T(x) and S(x) to determine the coefficients b and c, and ultimately calculate P(1). This involves polynomial division, a bit of algebraic manipulation, and a keen eye for detail. Now, why is understanding these relationships so important? Well, because the fact that P(x) divides both T(x) and S(x) means there's a connection, a common thread, that we can exploit. This shared factor gives us a system to work with, allowing us to set up equations and solve for our unknowns. It’s like having two different doors that both open with the same key - P(x) is that key!

So, before we jump into the calculations, let’s recap: We have three polynomials, P(x), T(x), and S(x). P(x) is a factor of both T(x) and S(x), and our goal is to find P(1). This means we need to figure out the coefficients of P(x) first. We will use the information about shared factors and polynomial division to get there. Keep in mind that polynomial factorization is not just a mathematical exercise; it's a fundamental tool in various fields, including engineering, computer science, and physics. Understanding how polynomials interact and share factors allows us to model and solve complex problems in these areas. In our specific scenario, knowing that P(x) is a factor of both T(x) and S(x) helps us constrain the possibilities for P(x). If P(x) were just a factor of T(x), there might be several possible quadratics that fit the bill. But the added condition of it also being a factor of S(x) significantly narrows down the options, making our task feasible. So, let's put on our thinking caps and get ready to dive into the mathematical journey ahead!

Finding a Common Factor

Okay, team, let's get our hands dirty and start crunching some numbers! Since P(x) is a factor of both T(x) and S(x), it means that any linear combination of T(x) and S(x) will also have P(x) as a factor. This is a neat trick that can help us simplify things. Think of it like this: if two numbers are divisible by the same factor, then any combination of those numbers (like adding or subtracting multiples of them) will also be divisible by that factor. In our case, the "numbers" are polynomials, and the "factor" is P(x).

So, let's create a new polynomial by taking a linear combination of T(x) and S(x). A clever way to do this is to eliminate the x⁴ term. Why eliminate x⁴? Because it simplifies the resulting polynomial, making it easier to work with. The fewer terms we have, the easier it is to find the factors. We can achieve this elimination by multiplying T(x) by 3 and subtracting S(x) from it. This gives us a new polynomial, which we'll call U(x):

U(x) = 3T(x) - S(x)

Let's plug in the expressions for T(x) and S(x):

U(x) = 3(x⁴ + 6x² + 25) - (3x⁴ + 4x² + 28x + 5)

Now, let's expand and simplify:

U(x) = 3x⁴ + 18x² + 75 - 3x⁴ - 4x² - 28x - 5

Notice how the 3x⁴ terms cancel out? That's exactly what we wanted! This leaves us with a simpler polynomial:

U(x) = 14x² - 28x + 70

This U(x) is still divisible by P(x), because it's a linear combination of T(x) and S(x), both of which are divisible by P(x). Now, to make things even simpler, we can factor out a common factor from U(x). Looking at the coefficients, we see that 14 is a common factor:

U(x) = 14(x² - 2x + 5)

So, we have U(x) expressed as 14 times another quadratic polynomial (x² - 2x + 5). Since 14 is just a constant, it doesn't affect the factors that contain x. Therefore, we can conclude that P(x) must be a factor of (x² - 2x + 5). This is a huge step forward! We've gone from dealing with quartic polynomials to a much simpler quadratic one. But why was this step so critical? It allowed us to narrow down the possibilities for P(x). Instead of considering all possible quadratic factors of two different quartic polynomials, we now only need to consider the factors of a single, simpler quadratic. This significantly reduces the complexity of the problem.

Therefore, we can now strongly suspect that P(x) is proportional to (x² - 2x + 5). We're not quite there yet, but we're getting closer to identifying P(x) and, ultimately, finding P(1). We need to confirm this suspicion and determine the exact form of P(x). Next, we'll use this information along with the original polynomials to nail down the values of b and c in P(x).

Identifying P(x)

Alright, detectives, we've cornered our suspect! We've deduced that P(x) is likely proportional to (x² - 2x + 5). But let's not jump to conclusions just yet. We need to confirm this and nail down the exact coefficients of P(x). Remember, P(x) has the form x² + bx + c. So, our mission now is to find the values of b and c.

From our previous step, we found that U(x) = 14(x² - 2x + 5). Since P(x) is a factor of U(x), and P(x) is a quadratic, it's highly probable that P(x) = x² - 2x + 5. Notice that the leading coefficient of x² in P(x) is 1, which matches the form given in the problem. This makes our hypothesis even stronger. However, we need to be absolutely sure. We can't just assume; we need to prove it.

So, how do we confirm this? We go back to our original conditions: P(x) must be a factor of both T(x) and S(x). Let's start with T(x) = x⁴ + 6x² + 25. If P(x) is indeed a factor of T(x), then dividing T(x) by P(x) should give us a clean result with no remainder. Time for some polynomial division!

Dividing T(x) by P(x), we get:

(x⁴ + 6x² + 25) / (x² - 2x + 5) = x² + 2x + 5

Voila! The division works out perfectly, leaving us with another quadratic polynomial (x² + 2x + 5) and no remainder. This strongly supports our hypothesis that P(x) = x² - 2x + 5. But we're not done yet! We have another condition to check: P(x) must also be a factor of S(x). So, let's divide S(x) by P(x):

S(x) = 3x⁴ + 4x² + 28x + 5

Dividing S(x) by P(x), we get:

(3x⁴ + 4x² + 28x + 5) / (x² - 2x + 5) = 3x² + 6x + 1

Another clean division! We get a quadratic polynomial (3x² + 6x + 1) with no remainder. This confirms beyond any reasonable doubt that P(x) = x² - 2x + 5 is indeed the correct quadratic factor. We've successfully identified P(x)! We now know the values of b and c. Comparing P(x) = x² - 2x + 5 with the general form P(x) = x² + bx + c, we see that:

  • b = -2
  • c = 5

We're on the home stretch now, guys! We've identified P(x), and we know its coefficients. All that's left is to answer the original question: What is P(1)?

Calculating P(1)

Okay, the moment we've all been waiting for! We've done the hard work of identifying P(x). Now, calculating P(1) is the easy part, like the cherry on top of a delicious mathematical sundae. We've established that:

P(x) = x² - 2x + 5

So, to find P(1), we simply substitute x = 1 into the expression for P(x). Get ready for some simple arithmetic!

P(1) = (1)² - 2(1) + 5

Let's break it down step by step:

  • (1)² = 1
  • -2(1) = -2

Now, plug these values back into the expression:

P(1) = 1 - 2 + 5

Adding and subtracting, we get:

P(1) = 4

And there you have it, folks! We've cracked the code. P(1) = 4. This is the solution to our problem. But let's take a moment to appreciate the journey we've been on. We started with a somewhat daunting problem involving polynomial factorization and ended with a simple numerical answer. We used clever techniques like finding linear combinations of polynomials and polynomial division to unravel the problem. This is what makes mathematics so fascinating – the ability to solve complex problems by breaking them down into smaller, manageable steps.

So, to recap, we were given that P(x) is a factor of both T(x) and S(x). We used this information to find a simpler polynomial U(x) that also has P(x) as a factor. This allowed us to identify P(x) as x² - 2x + 5. Finally, we substituted x = 1 into P(x) to find P(1) = 4. This problem beautifully illustrates the power of polynomial factorization and how it can be used to solve seemingly complex problems. The key takeaway here is the importance of understanding the relationships between polynomials and how shared factors can provide valuable clues. It's like a mathematical puzzle where each piece of information fits together to reveal the final answer.

Conclusion

So, there you have it! The value of P(1) is 4. We solved this problem by leveraging the properties of polynomial factors and using techniques like polynomial division and linear combinations. Remember, guys, math isn't just about getting the right answer; it's about the journey and the problem-solving skills you develop along the way. Keep practicing, keep exploring, and you'll be amazed at what you can achieve!